What is the integral of 1/(e^{2x)+e^{-2x}} ?

The integral of 1/(e^{2x)+e^{-2x}} is 1/2 arctan(e^{2x})+C

integral of 1/(e^{2x)+e^{-2x}}

\int \frac{1}{e ^{2 x} + e ^{- 2 x}} d x

\frac{1}{2} arctan (e^{2 x}) + C

Solution steps

\int \frac{1}{e ^{2 x} + e ^{- 2 x}} d x

Apply u-substitution

= \int \frac{1}{2 ( u ^{2} + 1 )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= \frac{1}{2} arctan (u)

Substitute back

u = e^{2 x}

= \frac{1}{2} arctan (e^{2 x})

Add a constant to the solution

= \frac{1}{2} arctan (e^{2 x}) + C

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What is the integral of 1/(e^{2x)+e^{-2x}} ?
The integral of 1/(e^{2x)+e^{-2x}} is 1/2 arctan(e^{2x})+C